Fix typos

finsberg · finsberg · commit ad336b943800 · 2025-02-15T23:40:11.000+01:00
diff --git a/examples/electro-mechanics/electro_mechanics.py b/examples/electro-mechanics/electro_mechanics.py
@@ -26,7 +26,7 @@
 # \nabla \cdot ((M_i + M_e) \nabla u_e) = - \nabla \cdot (M_i \nabla v)
 # $$
 #
-# Here the ordinary differential equations (ODEs), defined by $f$ describes the electro-chemical state of the muscle cells, characterized by the state vector $s$. Furthermore v is the transmembrane potential, $\lambda$ is the fiber stretch ratio and $\dot{\lambda} = \frac{\partial \lambda}{\partial t}$ is the stretch rate. The electrical state of the tissue is described by the Bi-domain model, which could also be reduced to the Mondomain model (see {cite}`sundnes2007computing` for more info). The function $I_{\text{ion}}$ describes the ionic current accross the cell membrane, while $M_i$ and $M_e$ are intracellular and extracellular tissue conductivities respectively.
+# Here the ordinary differential equations (ODEs), defined by $f$ describes the electro-chemical state of the muscle cells, characterized by the state vector $s$. Furthermore $v$ is the transmembrane potential, $\lambda$ is the fiber stretch ratio and $\dot{\lambda} = \frac{\partial \lambda}{\partial t}$ is the stretch rate. The electrical state of the tissue is described by the Bi-domain model, which could also be reduced to the Mondomain model (see {cite}`sundnes2007computing` for more info). The function $I_{\text{ion}}$ describes the ionic current accross the cell membrane, while $M_i$ and $M_e$ are intracellular and extracellular tissue conductivities respectively.
 #
 # The mechanical equilibrium are described by the following system of equations
 #
@@ -49,25 +49,25 @@
 # \mathbf{P}_a = \mathbf{P}_a(s, \lambda, \dot{\lambda}).
 # $$
 #
-# Here $\Psi$ is a strain energy function which relates stress to strain. For examples we can choose a NeoHookean material model (which is typically for rubber like material)
+# Here $\Psi$ is a strain energy function which relates stress to strain and depends on the material under consideration (This is often denoted as the "constitutive relation"). For examples we can choose a NeoHookean material model (which is typical for rubber-like materials)
 #
 # $$
 # \Psi(\mathbf{F}) = \frac{\mu}{2}( \mathrm{tr}(\mathbf{F}^T \mathbf{F}) - 3)
 # $$
 #
 # The active stress $\mathbf{P}_a$ depends on the electro-chemical state, and is typically described via a model for the crossbrigde dynamics.
 #
-# Now the mechanics system is typically solved for the displacement field $\mathbf{u}$, which relates coordinates in the reference configuration ($\mathbf{X}$) to coordinates in the current configuration ($\mathbf{x}$), by $\mathbf{u} = \mathbf{x} - \mathbf{X}$. We also have the deformation gradient $\mathbf{F} = \nabla \mathbf{u} + \mathbf{I}$. Furthermore the stretch is typically defined as the stretch along the fiber direction. If we denote the fiber orientation in the reference configuration as $\mathbf{f}_0$, then $\lambda = | \mathbf{F}\mathbf{f}_0 | =  | \mathbf{f} | $. For example one can consider a tissue slab with fibers oriented in the $x$-direction. In this case $\mathbf{f}_0 = (1 \; 0 \; 0)^T$
+# Now the mechanics system is typically solved for the displacement field $\mathbf{u}$, which relates coordinates in the reference configuration ($\mathbf{X}$) to coordinates in the current configuration ($\mathbf{x}$), by $\mathbf{u} = \mathbf{x} - \mathbf{X}$. We also have the deformation gradient $\mathbf{F} = \nabla \mathbf{u} + \mathbf{I}$. Furthermore the stretch is typically defined as the stretch along the fiber direction. If we denote the fiber orientation in the reference configuration as $\mathbf{f}_0$, then $\lambda = | \mathbf{F}\mathbf{f}_0 | =  | \mathbf{f} | $. For example one can consider a tissue slab with fibers oriented in the $x$-direction, in which case we would have $\mathbf{f}_0 = (1 \; 0 \; 0)^T$
 #
 # ## Reduction to 0D
 #
-# To reduce the problem to 0D, one need to consider a simplified experiment. For example if we consider a rectangle length $l_x$, heigth $l_y$ and depth $l_z$. Now imaging that we apply a deformation that changes the length from $l_x$ to $l_x + \Delta x = L_x$. We can quantify this deformation using the stretch
+# To reduce the problem to 0D, one need to consider a simplified experiment. For example we can consider a rectangle of length $l_x$, a heigth of $l_y$ and a depth of $l_z$. Now imaging that we apply a deformation that changes the length from $l_x$ to $l_x + \Delta x = L_x$. We can quantify this deformation using the stretch
 #
 # $$
 # \lambda = \frac{L_x}{l_x} = \frac{l_x + \Delta x}{l_x} = 1 + \frac{\Delta x}{l_x}
 # $$
 #
-# If we now assume the material is incompressible, and that the compression along the $y$- and $z$-axis are equal then we can write the following deformation gradient
+# If we now assume that the material is incompressible, and that the compression along the $y$- and $z$-axis are equal then we can write the following deformation gradient
 #
 # $$
 # \mathbf{F} = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda^{-1/2} & 0 \\ 0 & 0 & \lambda^{-1/2}\end{pmatrix}
@@ -152,7 +152,7 @@
 
 display(mech_model.first_piola_kirchhoff(experiment.F).simplify())
 
-# Note here that the variable $p$ is introduced. This variable acts as a Lagrange multiplier to enforce incompressibilty
+# Note here that the variable $p$ is introduced. This variable acts as a Lagrange multiplier to enforce incompressibility
 
 print(mech_model.compressibility)
 print(mech_model.compressibility.str())
@@ -165,7 +165,7 @@
 # \mathbf{P}_a = \mathbf{P}_a(s)
 # $$
 #
-# We will now update our model to use an active stress model, which contains a variable $T_a$ that describes the stregth of the active force and a fiber direction (which in our case is chosen to be along the $x$-axis)
+# We will now update our model to use an active stress model, which contains a variable $T_a$ that describes the strength of the active force and a fiber direction (which in our case is chosen to be along the $x$-axis)
 
 active_model = zero_mech.active.ActiveStress()
 display(active_model)
@@ -223,7 +223,7 @@
 mon = model["monitor_values"]
 # -
 
-# We set the initial $\lambda$ to 1.0 and $\dot{\lambda}$ to 0.0. Since we would need to update $\lambda$ and $\dot{\lambda}$ we need to also keep track of the previous value of $\lambda$.
+# We set the initial $\lambda$ to 1.0 and $\dot{\lambda}$ to 0.0. Since we would need to update $\lambda$ and $\dot{\lambda}$, we need to also keep track of the previous value of $\lambda$.
 
 # +
 lmbda_index = model["parameter_index"]("lmbda")
@@ -237,7 +237,7 @@
 
 # -
 
-# Now, we will create a function that will solve the mechanics problem (i.e P_xx = P_yy = 0) for $\lambda$ and $p$, given an active stress $T_a$
+# Now, we will create a function that will solve the mechanics problem (i.e $P_{xx} = P_{yy} = 0$) for $\lambda$ and $p$, given an active stress $T_a$
 
 def func(x, Ta):
     lmbda, p = x
@@ -254,7 +254,7 @@ def func(x, Ta):
     return np.array([P11, P22], dtype=np.float64)
 
 
-# Now let us loop over the time steps and in each iteration we first solve the EP problem by calling the `fgr` (forward generalized rush larsen) function, and then we solve the mechanics problem using a root-finding algorithm. An important observation to make here is that we do not update $\lambda$ or $\dot{\lambda}$ here (we have commented out the relevant lines that would update these values in the EP model)
+# Now let us loop over the time steps and in each iteration we first solve the EP problem by calling the `fgr` (forward generalized rush larsen) function, and then we solve the mechanics problem using a root-finding algorithm. An important observation to make here is that we do not update $\lambda$ or $\dot{\lambda}$ (we have commented out the relevant lines that would update these values in the EP model)
 
 # +
 from scipy.optimize import root
@@ -385,9 +385,9 @@ def func(x, Ta):
 plt.show()
 
 
-# Yikes!! That did not work well. We see that the solution starts oscillating. The reason this happens is because $T_a$ depends heavily on $\lambda$ and $\dot{\lambda}$, but we also have that $\lambda$ and $\dot{\lambda}$ depends heavily on $T_a$. For example if we stretch the material (i.e increase $\lambda$), then the [Frank Starling law](https://en.wikipedia.org/wiki/Frank%E2%80%93Starling_law) states that $T_a$ should be increased, but when $T_a$ increases we get more force that compress the material which will bring $\lambda$ down and it is not hard do see why this system can start oscillating.
+# Yikes!! That did not work well. We see that the solution starts oscillating. The reason this happens is because $T_a$ depends heavily on $\lambda$ and $\dot{\lambda}$, but we also have that $\lambda$ and $\dot{\lambda}$ depends heavily on $T_a$. For example if we stretch the material (i.e increase $\lambda$), then the [Frank Starling law](https://en.wikipedia.org/wiki/Frank%E2%80%93Starling_law) states that $T_a$ should be increased, but when $T_a$ increases we get more force that compress the material which will bring $\lambda$ down, and it is not hard do see why this system can start oscillating.
 #
-# The solution to the problem is to make sure we update $T_a$ whenever we solve the mechanics model, which means that we would need to run the ODE systen during every iteration whithin the root finding algorithm. In other words, let us update the function as follows
+# The solution to the problem is to make sure we update $T_a$ whenever we solve the mechanics model, which means that we would need to integrate the ODE systen during every iteration within the root finding algorithm. In other words, let us update the function as follows
 
 def func(x, y, ti, dt, params, new_y, prev_lmbda):
     lmbda, p = x
@@ -412,7 +412,7 @@ def func(x, y, ti, dt, params, new_y, prev_lmbda):
     return np.array([P11, P22], dtype=np.float64)
 
 
-# Note that whenever we now get a new guess for $\lambda$ (and $p$), we update $T_a$ accordingly. Also now that we also need to pass in the previous $\lambda$ (in order to compute $\dot{\lambda}$ as well as the previous state, the parameters and the time step. Our time loop will therefore need to be updated accordingly.
+# Note that whenever we now get a new guess for $\lambda$ (and $p$), we update $T_a$ accordingly. We now need to also pass the previous $\lambda$ (in order to compute $\dot{\lambda}$) as well as the previous state, the parameters and the time step. Our time loop will therefore need to be updated accordingly.
 
 # +
 lmbda_value = 1.0
